Question

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3 A four-pole, 60 Hz, 460 V, 5 HP induction motor has the following equivalent circuit parameters: Rs = 1.21 , Rr = 0.742, Xs = 3.102, Xr =2.41 , 100 RPM. Find the starting and no-load current for this machine. 4 A motor is rated 50 HP, 60 Hz, 460 V, four poles. When operating at rated conditions, efficiency of 89.6%, P.F =79.8%, s = 3%. If the motor is operated at 55 Hz, 430 V, and nr = 1605 RPM, what is the approximate developed torque?

that is all the info I have for that problem

Answer 1

0 Induction motor no.ob poly pry f = 60142 29 VH=1460V, terminal voltage power = 5hp RS= 1.21 XS = 3.100 Naluorome RY-0.742.0 Equivalent circuit- voro 1.21 jalan Meme 12.41 0.742 5 460 at no load starting sa I - 4601.13 (1.21+0.742) +j (310+ 2.41 ) Is 460153 1.952 +35.51

= 45.43 L470.49 Ampsit van Hook] starting Current = 45.43 270.49 Amps ☺ powers 5*7462 3130 rated or nominal value ho PM = 3730 3*460 Scotton hoone 4.6815 Amps • No-load current = (30 to 4o y of rated current = 0.40* 4.6815 =1.87 Amps Approximate

Officiency = 0.896 : ! Pf=79.8= 0, 798 lag Slip=s= 0.03 output power ilp power&n = 50*746* = 33420.8 whts out speed Nr= NS(1-5) = 120x60 * 11-0.03) - 4 (NS=120+) = 1796 rpone Ns=1204 = 1746 rpm output torque = out put power wr 334206 annr - 60 = 33420.8* 60 2*3.14 * 1746 = 182.87 N-N

full load torque depends w RO no-1뀜ort | 21. 21 품이 가히 = 101 성 더 . (19.32 - N - 1099